Solutions to Homework Assignment 1

1. (10pts) Solve the following first order PDE and find where the solution is defined in thex− y plane.

ux + xyuy = 0, u(x, 1) = x2

Solution: By the method of characteristics

dx

1=

dy

xy=

du

0

Since u(x, 1) = x2, we use y as parameter so we have

dx

dy=

1

xy, x(1) = ξ;

du

dy= 0, u(1) = ξ2

Solving the first ODE we get

x2 = 2 log y + C, x2 = 2 log y + ξ2

and so ξ2 = x2 − 2log(y)The solution is

u(x, y) = x2 − 2 log(y)

where y > 0 (since the initial data curve passes through y = 1).

2. (20pts) Solve xux + xyuy = u for u = u(x, y) with date u(1, y) = y2 for 0 ≤ y ≤ 1 andfind where the solution is defined in the x− y plane.Solution: By the method of characteristics

dx

x=

dy

xy=

du

u

Since u(1, y) = y2, we use x as parameter so we have

dy

dx=

xy

x, y(1) = ξ, 0 ≤ ξ ≤ 1;

du

dy= u, u(1) = ξ2

1

Solving the first ODE we gety = Cex, y = ξex−1

and so ξ = ye1−x

The solution isu(x, y) = (ye1−x)2

where 0 ≤ ξ = ye1−x ≤ 1.The domain of definition is 0 ≤ y ≤ ex−1.

3. (15pts) Solve the following first order PDE and find where the solution becomes un-bounded in the x− y plane.

x2ux + xyuy = u3, u = 1 on the curve y = x2

Solution: By the method of characteristics

dx

x2=

dy

xy=

du

u3

Since u = 1 on y = x2, we use s as parameter so we have

dx

ds= x2, x(0) = ξ,

dy

ds= xy, y(0) = ξ2,

du

ds= u3, u(0) = 1

Solving the first ODE we get

−1

x= s + C, x =

ξ

1− sξ

Solving the second ODE we get

log y = − log(1− sξ) + C, y =ξ2

1− sξ

Solving the last ODE we get

− 1

2u2= s + C, u2 =

1

1− 2s

2

Eliminating ξ and s we obtain ξ = yx, s = x2−y

xyand so

u2 =xy

xy − 2(x2 − y)

Since u = 1 on y = x2, we take

u =

√xy

xy − 2(x2 − y)

The blow up curve is xy − 2(x2 − y) = 0.The domain of definition is xy

xy−2(x2−y)> 0.

4. (20pts) Solve ut + t2ux = 4u for x > 0, t > 0 with u(0, t) = h(t) and u(x, 0) = 1.Solution: By the method of characteristics

dt

1=

dx

t2=

du

4u

There are two initial date curves so this problem can be decomposed into two systems of ODEs.ODE1: For u(x, 0) = 1, x > 0, we use t as parameter

dx

dt= t2, x(0) = ξ, ξ > 0;

du

dt= 4u, u(0) = 1

Solving the first ODE we get

x =1

3t3 + ξ

Solving the second ODE we getu = e4t

in the region ξ = x− 13t3 > 0.

ODE2: For u(0, t) = h(t), t > 0, we use x as parameter

dt

dx=

1

t2, t(0) = ξ, ξ > 0;

du

dx=

4u

t2, u(0) = h(ξ)

Solving the first ODE we gett = (3x + ξ)1/3

Solving the second ODE we get

log u = 4(3x + ξ)1/3 + C, u = h(ξ)e−4ξ1/3

e4(3x+ξ)1/3

3

and henceu = h(t3 − 3x)e−4(t3−3x)1/3

e4t

in the region ξ = t3 − 3x > 0.In conclusion, we have

u(x, t) =

{e4t, if x > t3

3

u = h(t3 − 3x)e−4(t3−3x)1/3e4t, if x < t3

3

5. (15pts) Solve xux + yuy = 2 with date u(x, 1) = x2 for −∞ < x < +∞. Explain why wecan not determine u(x, y) uniquely for y ≤ 0.Solution: By the method of characteristics

dx

x=

dy

y=

du

2

For u(x, 1) = x2, we use y as parameter

dx

dy=

x

y, x(1) = ξ;

du

dy=

2

y, u(1) = ξ2

Solving the first ODE we getx = ξy

Solving the second ODE we getu = 2 log y + ξ2

So the solution is

u = 2 log y +x2

y2

The domain of definition is y > 0, since log|y| is not defined at y = 0 and the initial data curvepasses through y > 0.

6.(20pts) Let u(x, y) solve the first order PDE

xux + yuy = xu

(a). Find the general solutions. (b) Suppose we put u = h(x) on y = x. Derive the conditionthat h(x) must satisfy for a solution to exist.

4

Solution: By the method of characteristics

dx

x=

dy

y=

du

xu

The characteristics are λ = xy.

(a). Changing variables

x′= x, λ =

x

y, u(x, y) = U(x

′, λ)

we havexUx′ = x

′U,Ux′ = U

and soU = Cex

′= f(λ)ex

′

The general solution is

u = f(x

y)ex

(b). When y = x, we haveu = f(1)ex = h(x)

and so h(x) must be of the formh(x) = Bex

where B is a constant.If h(x) = Bex then the solution u is

u = f(x

y)ex, f(1) = B

5